26.8 gm of Na₂SO₄.nH₂O contains 12.6 g of water. The value of n is;

First, separate the total mass of the hydrate into the mass of the anhydrous salt and the mass of the water of crystallisation.

$m\_\{\backslash text\{anhydrous\}\}\; =\; 26.8\backslash text\{\; g\}\; -\; 12.6\backslash text\{\; g\}\; =\; 14.2\backslash text\{\; g\}$

Next, calculate the molar mass of anhydrous sodium sulfate (Na₂SO₄).

$M\_\{\backslash text\{Na\}\_2\backslash text\{SO\}\_4\}=2(23)+32+4(16)=142\backslash text\{\; g\; mol\}^\{-1\}$

Find the number of moles of Na₂SO₄ present in the 14.2 g of anhydrous salt.

$n\_\{\backslash text\{salt\}\}\; =\; \backslash frac\{14.2\backslash text\{\; g\}\}\{142\backslash text\{\; g\; mol\}^\{-1\}\}\; =\; 0.10\backslash text\{\; mol\}$

The mass of one mole of water is approximately 18 g.

For a hydrate containing n molecules of water per formula unit, the total mass of water is:

$m\_\{\backslash text\{water\}\}\; =\; n\; \backslash times\; n\_\{\backslash text\{salt\}\}\; \backslash times\; 18\backslash text\{\; g\; mol\}^\{-1\}$

Set this equal to the given water mass (12.6 g) and solve for n.

$12.6\backslash text\{\; g\}\; =\; n\; \backslash times\; 0.10\backslash text\{\; mol\}\; \backslash times\; 18\backslash text\{\; g\; mol\}^\{-1\}$

$n\; =\; \backslash frac\{12.6\backslash text\{\; g\}\}\{0.10\backslash text\{\; mol\}\; \backslash times\; 18\backslash text\{\; g\; mol\}^\{-1\}\}\; =\; 7$

Therefore, the hydrate contains seven water molecules per formula unit.